Download Introduction, Binary representation - CSE-IITM

05/08/15
CS1101
Introduction to Programming
Instructors: Krishna Sivalingam, V Krishna Nandivada, Rajsekar M,
Co-ordinator: Madhu Mutyam
Course Material – SD, SB, PSK, NSN, DK, TAG – CS&E, IIT M
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Course Outline
•  Introduction to Computing
•  Programming (in C)
•  Exercises and examples from the mathematical
area of Numerical Methods
•  Problem solving using computers
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Evaluation
•  Two Quizzes – 30
•  Programming Assignments – 25
•  End of Semester Exam – 45
•  Attendance – taken in the lab and in lectures
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Class Hours
•  Class meets 3 times a week (E2 slot)
–  Monday 2.55 – 3.45 PM
–  Tuesday 1.00 – 1.50 PM
–  Wednesday 4.55 – 5.45 PM
•  Venue
–  CRC 102 / CRC 103
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Programming Assignments
•  Class split into batches; One batch per weekday
•  Time: 7:30 – 9:30 PM
•  Venue: Departmental Computing Facility (DCF)
in CSE Dept.
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Policies
•  Strictly no cell phone usage in class
–  Keep your cell phone turned off
•  Not even in silent mode
–  No SMS, chat etc.
•  Cell phones will be confiscated if there are
violations
–  Returned only after 2 weeks
•  Repeat violations
–  Students will be sent to the Dean (Acad)
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What is this CS110 about?
•  Computer and its components
•  Computing
•  Programming Languages
•  Problem Solving and Limitations of a Computer
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Common uses of a Computer
•  As a tool for storing and retrieving information
–  Extracting and storing information regarding students
entering IIT
•  As a tool for providing services to customers
–  Billing, banking, reservation
•  As a calculator capable of user-defined
operations
–  Designing electrical circuit layouts
–  Designing structures
–  Non-destructive testing and simulation
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What is a Computer?
•  A computer is a programmable machine
•  Its behavior is controlled by a program
•  Programs reside in the memory of the machine
–  “The stored program concept”
Charles Babbage
Von Neumann
Alan Turing
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Early Computing Hardware
The Slide rule
The Chinese Abacus
The gear replaced the beads in
early mechanical calculators
“History of computing hardware”
From Wikipedia, the free encyclopedia
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The Difference Engine : ~1850’s
Part of Babbage's difference
engine, assembled after his
death by Babbage's son, using
parts found in his laboratory.
The London Science Museum's
replica Difference Engine, built
from Babbage's design.
An automatic, mechanical
calculator designed to tabulate
polynomial functions
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The First Programmer
Augusta Ada King, Countess of
Lovelace (December 10, 1815 –
November 27, 1852), born Augusta
Ada Byron, is mainly known for
having written a description of
Charles Babbage's early mechanical
general-purpose computer, the
analytical engine.
The programming language ADA is named after her.
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ENIAC – The First Electronic Computer
Physically, ENIAC was
massive compared to modern
PC standards. It contained
17,468 vacuum tubes, 7,200
crystal diodes, 1,500 relays,
70,000 resistors, 10,000
capacitors and around 5
million hand-soldered joints. It
weighed 27 tons, was roughly
2.4 m by 0.9 m by 30 m, took
up 167 m², and consumed 150
kW of power.
Electronic Numerical Integrator and
Computer, 1946-55 (Univ. Penn.)
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Core i7 Processor (desktop version)
2008-15: Intel Core i7 Processor
Clock speed: >2.5 GHz
No. of Transistors: 0.731-1.3B
Technology: 45-22nm CMOS
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Area: 263-181mm2
The Computing Machine
PROCESSOR
MEMORY
01234…….
(say) 256 MEGABYTES
The computer is made up of a processor and a
memory. The memory can be thought of as a series
of locations to store information.
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The Computing Machine
PROCESSOR
MEMORY
01234…….
program
256 MEGABYTES
data
•  A program is a sequence of instructions assembled
for some given task
•  Most instructions operate on data
•  Some instructions control the flow of the operations
•  It is even possible to treat programs as data. By doing
so a program could even modify itself.
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Variables
•  Data is represented as binary strings
–  It is a sequence of 0’s and 1’s (bits), of a
predetermined size – “word”. A byte is made of 8 bits.
•  Each memory location may be given a name.
•  The name is the variable that refers to the data
stored in that location
–  e.g. rollNo, classSize
•  Variables have types that define the interpretation
of data
–  e.g. integers (1, 14, 25649), or characters (a, f, G, H)
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Instructions
•  Instructions take data stored in variables as
arguments
•  Some instructions do some operation on the data
and store it back in some variable
–  e.g. The instruction “X←X+1” on integer type says
that “Take the integer stored in X, add 1 to it, and
store it back in (location) X”
•  Other instructions tell the processor to do
something
–  e.g. “jump” to a particular instruction next, or to exit
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Programs
•  A program is a sequence of instructions
•  Normally the processor works as follows,
–  Step A: pick next instruction in the sequence
–  Step B: get data for the instruction to operate upon
–  Step C: execute instruction on data (or “jump”)
–  Step D: store results in designated location (variable)
–  Step E: go to Step A
•  Such programs are known as imperative
programs
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Programming Paradigms
•  Imperative programs are sequences of instructions. They
are abstractions of how the von Neumann machine
operates
•  Pascal, C, Fortran
•  Object Oriented Programming Systems (OOPS) model
the domain into objects and interactions between them
•  Simula, CLOS, C++, Java
•  Logic programs use logical inference as the basis of
computation
•  Prolog
•  Functional programs take a mathematical approach of
functions
•  LISP, ML, Haskell
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A Limitation – Computer Arithmetic
•  Number of digits that can be stored is limited
•  Causes serious problems
Consider a computer that can store:
Sign, 3 digits and a decimal point
Sign and decimal point are optional
example : 212, -212, -21.2, -2.12, -.212
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More Examples
•  113. + -111. = 2.00
•  2.00 + 7.51 = 9.51
•  -111. + 7.51 = -103.49 (exact arithmetic)
But our computer can store only 3 digits.
So it rounds –103.49 to –103
This is a very important thing to know as a system
designer. Why?
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Why?
Consider 113. + -111. + 7.51
To us addition is associative
(a+b)+c = a+(b+c)
(113. + -111.) + 7.51 = 2.00 + 7.51 = 9.51
113. + (-111. + 7.51) = 113. – 103. = 10.0
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Conclusion
•  Computer is fast but restricted
•  So we must learn to use its speed
•  And manage its restrictions
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Books
•  Paul Deitel and Harvey Deitel. C: How to
Program.
•  V. Rajaraman: Computer Programming in C
•  R. G. Dromey: How to Solve It By Computer
•  Kernighan and Ritchie: The C Programming
Language
•  Kernighan and Pike: The Unix Programming
Environment
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Building Blocks of a Computer
Central
Processing
Unit
Input
Control Unit
Memory
ALU
Output
System Bus
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The Blocks, Their Functions
•  Input unit
–  Takes inputs from the external world via variety of
input devices – keyboard, mouse, etc.
•  Output Unit
–  Sends information (after retrieving, processing) to
output devices – monitors/displays, projectors, audio
devices, etc.
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More (try more filename on your Unix/Linux machine)
•  Memory
–  Place where information is stored
–  Primary memory
•  Electronic devices, used primarily for temporary storage
•  Characterized by their speedy response
–  Secondary Memory
•  Devices for long-term storage
•  Contained well tuned mechanical components, magnetic
storage media – floppies, hard disks
•  Compact Disks use optical technology
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Some More
(Commands are in /bin, /usr/bin. Use ls)
•  System Bus
–  Essentially a set of wires, used by the other units to
communicate with each other
–  transfers data at a very high rate
•  ALU – Arithmetic and Logic Unit
–  Processes data - add, subtract, multiply, …
–  Decides – after comparing with another value, for
example
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Finally (check man cp, man mv, man ls, man –k search string)
•  Control Unit
–  Controls the interaction among other units
–  Knows each unit by its name, responds to requests
fairly, reacts quickly on certain critical events
–  Gives up control periodically in the interest of the
system
Control Unit + ALU is called the CPU
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The CPU (editors vi, emacs used to create text)
• 
• 
• 
• 
• 
Can fetch an instruction from memory
Execute the instruction
Store the result in memory
A program – a set of instructions
An instruction has the following structure
Operation operands destination
•  A simple operation
add a, b
Adds the contents of memory locations a and b
and stores the result in location a
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Compilers
Human friendly languages à source code
Source code in a
Higher Level Language 1
Source code in a
Higher Level Language n
Compiler
Compiler
Assembly language code
Assembler, linker, loader
machine language code
Machine understandable language
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Assembly language
•  An x86/IA-32 processor can execute the
following binary instruction as expressed in
machine language:
Binary: 10110000 01100001
mov al,
061h
–  Move the hexadecimal value 61 (97 decimal) into the
processor register named ”al".
–  Assembly language representation is easier to
remember (mnemonic)
From Wikipedia
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Higher Level Languages
•  Higher level statement = many assembly
instructions
•  For example “X = Y + Z” could require the
following sequence
–  Fetch the contents of Y into R1
–  Fetch the contents of Z into R2
–  Add contents of R1 and R2 and store it in R1
–  Move contents of R1 into location named X
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Data Representation
•  Integers – Fixed Point Numbers
Decimal System - Base 10
(396)10 = (6 ×
100)
+ (9 ×
uses 0,1,2,…,9
101)
Binary System - Base 2
20)+(0
(11001)2 = (1 ×
= (25)10
×
+ (3 × 102) = (396)10
uses 0,1
21)+(0
×
22)+(1
× 23)+(1 × 24)
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Decimal to Binary Conversion
Convert (39)10 to binary form
Base = 2
2
2
2
2
39
19 + Remainder 1
9 + Remainder 1
4 + Remainder 1
2
2
2 + Remainder 0
1 + Remainder 0
0 + Remainder 1
39 = 2*19 + 1
= 2*(2*9 +1) + 1
= 22*9 + 21*1 + 1
= 22*(2*4+1) + 21*1 + 1
= 23*4+22*1+ 21*1 + 1
= 23*(2*2+0)+22*1+ 21*1 + 1
= 24*2+23*0+ 22*1+ 21*1 + 1
= 24*(2*1+0) + …
= 25*1+24*0+23*0+22*1+ 21*1+ 1
Put the remainders in reverse order
(100111)2 = (1 × 20 )+(1 × 21)+(1 × 22)+(0 × 23)+(0 × 24)+(1 × 25)
= (39)10
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Largest number that can be stored in m-digits
base - 10 :
(99999…9) = 10m - 1
base - 2 :
(11111…1) = 2m - 1
m=3
(999) = 103 - 1
(111) = 23 - 1
Limitation: Memory cells consist of 8 bits (1 byte)
multiples, each position containing 1 binary digit
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Sign - Magnitude Notation
Common cell lengths for integers : k = 16 or 32 or 64 bits
First bit is used for a sign
0 – positive number
1 – negative number
The remaining bits are used to store the binary
magnitude of the number.
Limit of 16 bit cell : (32,767)10 = ( 215 – 1)10
Limit of 32 bit cell : (2,147, 483,647)10 = ( 231 – 1)10
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One’s Complement Notation
In the one’s complement method, the negative of integer n
is
represented as the bit complement of binary n
E.g. : One’s Complement of (3)10 in a 3 - bit cell
complement of 011 : 100
-3 is represented as = (100)2
Arithmetic requires care:
2 + (-3) = 010 + 100 = 110 – ok
But, 3 + (-2) = 011 + 101 = 000 and carry of 1
need to add back the carry to get 001!
NOT WIDELY USED
000 :
001 :
010 :
011 :
100 :
101 :
110 :
111 :
0
+1
+2
+3
-3
-2
-1
-0
Zero has two
representations!
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Two’s Complement Notation
In the two’s complement method, the negative of integer n
in a k - bit cell is represented as 2k – n
Two’s Complement of n = (2k – n)
E.g. : Two’s Complement of (3)10 in a 3 - bit cell
-3 is represented as (23 - 3)10 = (5)10 = (101)2
000 :
001 :
010 :
011 :
100 :
101 :
110 :
111 :
Arithmetic requires no special care:
2 + (-3) = 010 + 101 = 111 – ok
3 + (-2) = 011 + 110 = 001 and carry of 1
we can ignore the carry!
WIDELY USED METHOD for –ve numbers
0
+1
+2
+3
-4
-3
-2
-1
(8 – 4)
(8 – 3)
(8 – 2)
(8 – 1)
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Two’s Complement Notation
The Two’s Complement notation admits one more
negative number than the sign - magnitude notation.
To get back n, read off the sign from the MSB
000 : 0
If –ve, to get magnitude, complement the cell and 001 : +1
010 : +2
add 1 to it!
011 : +3
100 : -4
E.g.: 101 à 010 à 011 = (-3)10
101 : -3
110 : -2
111 : -1
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Numbers with Fractions
Integer Part + Fractional Part
Decimal System - base 10
235 . 7846
Binary System - base 2
10011 . 11101 = (19.90625) 10
Fractional Part (0.7846)10
Fractional Part (0.11101)2
=
=
1
2
7
10
+
+
8
10 2
1
22
+
+
4
103
1
23
+
+
6
10 4
0
24
+
1
25
= 0.90625
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Binary Fraction → Decimal Fraction
(10.11)2
Integer Part (10)2
= 1*21 + 0*20 = 2
Fractional Part (11)2 =
Decimal Fraction = ( 2.75 )10
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Decimal Fraction → Binary Fraction (1)
Convert (0.90625)10 to binary fraction
0.90625
×2
0.8125 + integer part 1
×2
0.625
+ integer part 1
×2
0.25
+ integer part 1
×2
0.5
+ integer part 0
×2
0
0.90625 = ½(1+0.8125)
= ½(1+ ½(1+0.625))
= ½(1+ ½(1+ ½(1+0.25)))
= ½(1+½(1+ ½(1+½(0+0.5))))
= ½(1+½(1+½(1+½(0+½(1+0.0)))))
= ½+1/22+1/23+0/24 +1/25
= (0.11101)2
+ integer part 1
Thus, (0.90625)10 = (0.11101)2
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Decimal Fraction → Binary Fraction (2)
Convert (0.9)10 to binary fraction
0.9
×2
0.8
×2
0.6
×2
0.2
×2
0.4
×2
For some fractions, we do
not get 0.0 at any stage!
+ integer part 1 These fractions require an
infinite number of bits!
+ integer part 1
Cannot be represented
exactly!
+ integer part 1
+ integer part 0
0.8 + integer part 0
Repetition
(0.9)10 = 0.11100110011001100 . . . = 0.11100
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Fixed Versus Floating Point Numbers
Fixed Point: position of the radix point fixed
and is same for all numbers
E.g.: With 3 digits after decimal point:
0.120 * 0.120 = 0.014
A digit is lost!!
Floating point numbers: radix point can float
1.20 × 10-1 * 1.20 × 10-1 = 1.44 × 10-2
Floating point system allows a much wider range of
values to be represented
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Scientific Notation (Decimal)
0.0000747 = 7.47 × 10-5
31.4159265 = 3.14159265 × 101
9,700,000,000 = 9.7 × 109
Binary
(10.01)2 = (1.001)2 × 21
(0.110)2 = (1.10)2 × 2-1
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Using Floating Point Notation
For any number x
x ≈ ± q × 2n
32 bits :
First bit for sign
q – mantissa
n – exponent
(-39.9)10 = (-100111.1 1100)2
Next 8 bits for exponent
23 bits for mantissa
= -39. 90000152587890625
= (-1.001111 1100)2 × 25
Decimal Value of stored number (-39.9)10
= (-1. 001111 1100 1100 1100 11001) × 25
23 bit
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